(1-m^2)x^2+2(1-m)x-1=0
两个实数根,判别式≧0
[2(1-m)]^2-4(1-m^2)(-1)≧0
1-m≧0
m≦1,又已知m是非负整数,m≧0,则m=0或1
但是当m=1带回原式,不合,故m=0
若m是非负整数,且关于x的一元二次方程(1-m平方)x平方+2(1-m)x-1=0有两个实数根,求m的值
若m是非负整数,且关于x的一元二次方程(1-m平方)x平方+2(1-m)x-1=0有两个实数根,求m的值
RT,坐等
RT,坐等
数学人气:176 ℃时间:2019-09-30 07:16:10
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